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Chapter 4: Problem 18

Limits Evaluate the following limits. Use l'H贸pital's Rule when it is comvenient and applicable. $$\lim _{x \rightarrow-1} \frac{x^{4}+x^{3}+2 x+2}{x+1}$$

Short Answer

Expert verified
Question: Evaluate the limit using L'H么pital's Rule when applicable: $\lim_{x \rightarrow -1}\frac{x^4 + x^3 + 2x + 2}{x + 1}$. Answer: 1

Step by step solution

01

Check the conditions for L'H么pital's Rule

First, let's check the conditions for L'H么pital's Rule by plugging in the limit point (-1) into the function: $$\frac{(-1)^4 + (-1)^3 + 2(-1) + 2}{-1 + 1} = \frac{1 - 1 - 2 + 2}{0} =\frac{0}{0}$$ Since we get a 0/0 indeterminate form, we can apply L'H么pital's Rule in this case.
02

Apply L'H么pital's Rule by finding the derivatives of the numerator and the denominator

To apply L'H么pital's Rule, we need to find the derivatives of the numerator and the denominator with respect to x: Numerator derivative: $$\frac{d}{dx}(x^{4}+x^{3}+2x+2) = 4x^3 + 3x^2 + 2$$ Denominator derivative: $$\frac{d}{dx}(x+1) = 1$$ Now, we can rewrite our limit expression as the limit of the derivatives: $$\lim_{x \rightarrow -1} \frac{4x^3 + 3x^2 + 2}{1}$$
03

Evaluate the limit of the derivatives

To evaluate the limit, plug in the limit point (-1) into the expression: $$\frac{4(-1)^3 + 3(-1)^2 + 2}{1} =\frac{-4 + 3 + 2}{1} = 1$$
04

State the final answer

Since the limit of the derivatives is equal to 1, we can conclude that the original limit is: $$\lim _{x \rightarrow-1} \frac{x^{4}+x^{3}+2 x+2}{x+1} = 1$$

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